integral from 0 to 8 of sqrt(1+9/4 x)

Lösung

\int_{0}^{8} 1 + \frac{9}{4} x d x

\frac{4 ( 38 19 - 2 )}{27}

Dezimale

24.24269 \dots

Schritte zur Lösung

\int_{0}^{8} 1 + \frac{9}{4} x d x

Wende U-Substitution an

= \int_{1}^{19} \frac{4 u}{9} d u

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{4}{9} \cdot \int_{1}^{19} u d u

Wende die Potenzregel an

= \frac{4}{9} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{19}

Berechne die Grenzen:

\frac{38 19}{3} - \frac{2}{3}

= \frac{4}{9} (\frac{38 19}{3} - \frac{2}{3})

Vereinfache

= \frac{4 ( 38 19 - 2 )}{27}